class Solution {
public:
    int myAtoi(string str) {
        int sign=0;
        int firstNonSpace;
        int lastNonSpace;
        
        for(firstNonSpace=0;firstNonSpace<str.size();firstNonSpace++){
            if(str[firstNonSpace]!=‘ ‘){
                break;
            }
        }
        
 
if(firstNonSpace==str.size()){
            return 0;
        }
        else{
            if(str[firstNonSpace]==‘+‘){
                firstNonSpace++;
            }
            else if(str[firstNonSpace]==‘-‘){
                sign=1;
                firstNonSpace++;
            }
            if(firstNonSpace==str.size()){
                
 
return 0;
            }
            if((str[firstNonSpace]<‘0‘)||(str[firstNonSpace]>‘9‘)){
               return 0; 
            }
            
            for(lastNonSpace=firstNonSpace;lastNonSpace<str.size();){
                if((str[lastNonSpace]>=‘0‘)&&(str[lastNonSpace]<=‘
 
9‘)){
                    lastNonSpace++;
                }
                else{
                    break;
                }
            }
            if(lastNonSpace-firstNonSpace>10){
                if(sign==0){
                    return INT_MAX;
                }
                else{
                    return INT_MIN;
                }
            }
            if(lastNonSpace-firstNonSpace==10){
                if(sign==0&&str.substr(firstNonSpace,lastNonSpace-firstNonSpace)>="2147483647"){
                    return INT_MAX;
                }
                else if(sign==1&&str.substr(firstNonSpace,lastNonSpace-firstNonSpace)>="2147483648"){
                    return INT_MIN;
                }
            }
            int result=0;
                for(int i=firstNonSpace;i<lastNonSpace;i++){
                    result=(str[i]-‘0‘)+result*10;
                }
                if(sign==0){
                    return result;
                }
            else{
                return -result;
            }
        }
    }
};

擊敗30%。

然後我今天又交了一次，沒過。

" 0000000000012345678"

別人的代碼：

class Solution {
public:
    int myAtoi(string str) {
        if (str.empty()) return 0;
        int sign = 1, base = 0, i = 0, n = str.size();
        while (i < n && str[i] == ‘ ‘) ++i;
        if (str[i] == ‘+‘ || str[i] == ‘-‘) {
            sign = (str[i++] == ‘+‘) ? 1 : -1;
        }
        while (i < n && str[i] >= ‘0‘ && str[i] <= ‘9‘) {
            if (base > INT_MAX / 10 || (base == INT_MAX / 10 && str[i] - ‘0‘ > 7)) {
                return (sign == 1) ? INT_MAX : INT_MIN;
            }
            base = 10 * base + (str[i++] - ‘0‘);
        }
        return base * sign;
    }
};

來自：http://www.cnblogs.com/grandyang/p/4125537.html

leetcode 8 字符串轉整數